## anonymous 3 years ago how to differentiate f(x) = ln ( 1 + √ x / 1 - √ x )

1. anonymous

In general, $\frac{d}{dx} \ln( f(x) ) = \frac{1}{f(x)}\cdot f'(x)$

2. anonymous

how do i solve it? i cant seem to get the correct answer.

3. anonymous

Hint: You will have to use the chain rule by finding the derivative of the argument and multiplying it by the derivative that results without the chain rule. Are you aware of how to use the Chain Rule? If not, then I'll provide a solution.

4. anonymous

im aware of it. but how do i calculate it?

5. anonymous

The derivative of the argument, you mean?

6. anonymous

how do i apply chain rule in this equation

7. anonymous

Rewrite f(x): $$\Large f(x)=ln\frac{1+\sqrt x}{1-\sqrt x}=ln(1+\sqrt x)-ln(1-\sqrt x)$$ now just take the derivatives of each ln separately....

8. anonymous

Would you mind showing us your work? What have you found for the derivative of the argument of the natural log?

9. anonymous

dy/dx = [ (1/1+√ x) (+ 1/2√x) ] - [ (1/1-√x) - ( 1/2√ x) ]